Momentum And Collisions

Question

A ball of mass 0.120 kg is dropped from rest from a height of 1.25 m. It rebounds from the floor to reach a height of 0.600 m. What impulse was given to the ball by the floor?

kg·m/s

Answer 1

Hi Dominic!

 

The key here is that the impulse delivered by the floor to the ball equals the ball's change in momentum.  To determine the ball's change in momentum, we have to determine the difference between its velocity upon hitting the floor, and its velocity upon rebounding.  We could use kinematics to do this, but conservation of (mechanical) energy expedites the process.

 

For the time the fall is traveling in the air with only gravity doing work (air resistance ignored), mechanical energy is conserved.  At the highest point on each leg (when being dropped on the way down, and at maximum return height on the way back up), the ball is momentarily at rest, and all its mechanical energy is (gravitational) potential energy.  At the floor, all its mechanical energy is kinetic.  By conservation of mechanical energy, the kinetic energy it has at the floor must equal the potential energy it has at its maximum height.  Thus, for each leg (drop and return bounce), we can write:

 

mgh = (1/2)mv² --> v = √(2gh)

 

where 

 

h = maximum height

v = speed at the floor

 

Thus, its speed right before hitting the floor on the drop is:

 

v_o = √[2(9.8 m/s²)(1.25 m)] = 4.95 m/s

 

and its speed just after leaving the floor on its return bounce is:

 

v_f = √[2(9.8 m/s²)(0.6 m)] = 3.43 m/s

 

So, if we call the downward direction negative, the ball's initial velocity before hitting the floor is v_o = -4.95 m/s and its final velocity v_f = 3.43 m/s. 

 

The change in momentum, Δp , that the ball undergoes is thus:

 

_{Δp =} p_f - p_o = mv_f - mv_o = m(v_f - v_o) = (0.120 kg)(3.43 m/s - (-4.95 m/s) = (0.120)(8.38) = 1.00 kg·m/s.

 

As mentioned above, this change in momentum is -- by the impulse-momentum theorem -- identical to the impulse delivered to the ball by the floor.

 

I hope this helps!  Let me know if you have any further questions about this.